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Equilibrium - Acids, Bases and Salts

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Arrhenius Concept: Acids are substances that dissociate in water to give hydrogen ions H+(aq)H^+(aq), and bases are substances that produce hydroxyl ions OH−(aq)OH^-(aq).

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Brönsted-Lowry Theory: An acid is a proton (H+H^+) donor and a base is a proton acceptor. This leads to the concept of conjugate acid-base pairs which differ by a single proton.

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Lewis Concept: An acid is an electron pair acceptor (e.g., BF3BF_3, AlCl3AlCl_3) and a base is an electron pair donor (e.g., NH3NH_3, H2OH_2O).

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Ionic Product of Water (KwK_w): The product of concentrations of hydrogen and hydroxyl ions. At 298 K298\text{ K}, Kw=[H+][OH−]=1.0×10−14K_w = [H^+][OH^-] = 1.0 \times 10^{-14}.

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pH Scale: Defined as the negative logarithm (to the base 10) of the activity of hydrogen ions. pH=−log⁡[H+]pH = -\log[H^+].

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Ionization Constants: For a weak acid HAHA, Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]}. For a weak base BOHBOH, Kb=[B+][OH−][BOH]K_b = \frac{[B^+][OH^-]}{[BOH]}.

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Common Ion Effect: The suppression of the degree of dissociation of a weak electrolyte by the addition of a strong electrolyte containing a common ion.

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Buffer Solutions: Solutions which resist change in pHpH on dilution or with the addition of small amounts of acid or alkali. Types include Acidic Buffers (weak acid + salt with strong base) and Basic Buffers (weak base + salt with strong acid).

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Solubility Product (KspK_{sp}): The product of the molar concentrations of the constituent ions raised to the power of their stoichiometric coefficients in a saturated solution of a sparingly soluble salt.

📐Formulae

pH=−log⁡10[H3O+]pH = -\log_{10}[H_3O^+]

pOH=−log⁡10[OH−]pOH = -\log_{10}[OH^-]

pKw=pH+pOH=14 (at 298 K)pK_w = pH + pOH = 14 \text{ (at 298 K)}

Ka×Kb=KwK_a \times K_b = K_w

α=KaC (Ostwald’s Dilution Law for weak electrolytes)\alpha = \sqrt{\frac{K_a}{C}} \text{ (Ostwald's Dilution Law for weak electrolytes)}

pH=pKa+log⁡[Salt][Acid] (Henderson-Hasselbalch Equation for Acidic Buffer)pH = pK_a + \log\frac{[Salt]}{[Acid]} \text{ (Henderson-Hasselbalch Equation for Acidic Buffer)}

pOH=pKb+log⁡[Salt][Base] (Henderson-Hasselbalch Equation for Basic Buffer)pOH = pK_b + \log\frac{[Salt]}{[Base]} \text{ (Henderson-Hasselbalch Equation for Basic Buffer)}

Ksp=[Ay+]x[Bx−]y for salt AxByK_{sp} = [A^{y+}]^x [B^{x-}]^y \text{ for salt } A_x B_y

💡Examples

Problem 1:

Calculate the pHpH of a 0.10 M0.10\text{ M} solution of acetic acid (CH3COOHCH_3COOH). The ionization constant KaK_a for CH3COOHCH_3COOH is 1.8×10−51.8 \times 10^{-5}.

Solution:

For a weak acid, [H+]=Ka⋅C[H^+] = \sqrt{K_a \cdot C}. Substituting the values: [H+]=(1.8×10−5)×0.10=1.8×10−6≈1.34×10−3 M[H^+] = \sqrt{(1.8 \times 10^{-5}) \times 0.10} = \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3}\text{ M}. Now, pH=−log⁡[H+]=−log⁡(1.34×10−3)=3−log⁡(1.34)≈2.87pH = -\log[H^+] = -\log(1.34 \times 10^{-3}) = 3 - \log(1.34) \approx 2.87.

Explanation:

Since acetic acid is a weak acid, we use the approximation derived from Ostwald's Dilution Law to find the concentration of hydrogen ions before calculating the pHpH.

Problem 2:

The solubility of Ag2CrO4Ag_2CrO_4 is 1.3×10−4 mol L−11.3 \times 10^{-4}\text{ mol L}^{-1}. Calculate its solubility product (KspK_{sp}).

Solution:

The dissociation is Ag2CrO4(s)⇌2Ag+(aq)+CrO42−(aq)Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq). If solubility is SS, then [Ag+]=2S[Ag^+] = 2S and [CrO42−]=S[CrO_4^{2-}] = S. Ksp=[Ag+]2[CrO42−]=(2S)2(S)=4S3K_{sp} = [Ag^+]^2 [CrO_4^{2-}] = (2S)^2(S) = 4S^3. Substituting S=1.3×10−4S = 1.3 \times 10^{-4}: Ksp=4×(1.3×10−4)3=4×2.197×10−12=8.788×10−12K_{sp} = 4 \times (1.3 \times 10^{-4})^3 = 4 \times 2.197 \times 10^{-12} = 8.788 \times 10^{-12}.

Explanation:

The KspK_{sp} is calculated by establishing the relationship between molar solubility (SS) and the ions produced based on the stoichiometry of the salt.

Acids, Bases and Salts - Revision Notes & Key Formulas | CBSE Class 11 Chemistry